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Bài 1:
b: \(=\left(x-2y\right)\left(x+2y\right)+4\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y+4\right)\)
c: \(=\left(x+y-3\right)\left(x+y+3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: \(3xy^2-12x=3x\left(y^2-4\right)=3x\left(y-2\right)\left(y+2\right)\)
b: \(x^2-4y^2+4x+8y\)
\(=\left(x-2y\right)\left(x+2y\right)+4\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y+4\right)\)
Giúp với ạ
CM biểu thức sau không phụ thuộc vào x:
A= (2x+1) (x^2-x+2) - (2x-1) (x^2-2) - 7x+5
Cảm ơn ạ
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(2x+1\right)\left(x^2-x+2\right)-\left(2x-1\right)\left(x^2-2\right)-7x+5\)
\(=2x\left(x^2-x+2\right)+x^2-x+2-2x\left(x^2-2\right)+\left(x^2-2\right)-7x+5\)
\(=2x^3-2x^2+4x+x^2-x+2-2x^3+4x+x^2-2-7x+5\)
\(=5\)
Vậy biểu thức k phụ thuộc vào x
P/s: Ủng hộ nha
\(\left(2x+1\right)\left(x^2-x+2\right)-\left(2x-1\right)\left(x^2-2\right)-7x+5\)
\(=2x^3-2x^2+4x+x^2-x+2-\left(2x^3-4x-x^2+2\right)\)\(-7x+5\)\(=2x^3-2x^2+4x+x^2-x+2-2x^3+4x+x^2-2-7x+5\)
\(=5\)không phụ thuộc vào x
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(\left(x-3\right)^2-x\left(x+5\right)=9\)
\(\Leftrightarrow x^2-6x+9-x^2-5x=9\)
\(\Leftrightarrow x=0\)
b: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`(x+3)(x^2-5x+8)=(x+3).x^2`
`<=>(x+3)(x^2-5x+8-x^2)=0`
`<=>(x+3)(8-5x)=0`
`<=>` \(\left[ \begin{array}{l}x+3=0\\8-5x=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac85\\x=-3\end{array} \right.\)
Vậy `S={-3,8/5}`
`(x+3)(x^2-5x+8)=(x+3).x^2`
`<=>(x+3)(x^2-5x+8-x^2)=0`
`<=>(x+3)(-5x+8)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\-5x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{8}{5}\end{matrix}\right.\)
Vậy `S={-3;8/5}`.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(\left(x^2+4\right)^2-16x^2\)
\(=\left(x^2-4x+4\right)\left(x^2+4x+4\right)\)
\(=\left(x-2\right)^2\cdot\left(x+2\right)^2\)
c: \(x^5-x^4+x^3-x^2\)
\(=x^4\left(x-1\right)+x^2\left(x-1\right)\)
\(=x^2\left(x-1\right)\left(x^2+1\right)\)
Lời giải:
a. Bạn xem lại đề
b. \((x^2+4)^2-16x^2=(x^2+4)^2-(4x)^2=(x^2+4-4x)(x^2+4+4x)\)
\(=(x-2)^2(x+2)^2\)
c.
\(x^5-x^4+x^3-x^2=x^4(x-1)+x^2(x-1)=(x^4+x^2)(x-1)\)
\(=x^2(x^2+1)(x-1)\)
a) \(x^3-x^2-12x=x\left(x^2-x-12\right)=x\left(x-4\right)\left(x+3\right)\)
b) \(x^2+7x=x\left(x+7\right)\)
\(a,=x\left(x^2-x-12\right)=x\left(x^2-4x+3x-12\right)\\ =x\left(x-4\right)\left(x+3\right)\\ b,=x\left(x+7\right)\)